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Find the limit of the equation $$\lim_{x\to-\infty} (x+\sqrt{x^2 + 2x})$$

I start by multiplying with the conjugate:

$$\lim_{x\to-\infty} \left[(x+\sqrt{x^2 + 2x})\left({x - \sqrt{x^2 + 2x}\over x - \sqrt{x^2+2x}}\right)\right]$$

$$\lim_{x\to-\infty} {x^2 - (x^2 + 2x)\over x - \sqrt{x^2+2x}}$$

$$\lim_{x\to-\infty} {-2x\over x - \sqrt{x^2+2x}}$$

divide by highest power of denominator

$$\lim_{x\to-\infty} {(\frac1x)(-2x)\over (\frac1x) x - ({1\over \sqrt{x^2}})\sqrt{x^2+2x}}$$

$$\lim_{x\to-\infty} {-2\over 1 - \sqrt{1+\frac2x}} = {-2 \over 1-\sqrt{1 + 0}} = {-2 \over 0}$$ but I know this is wrong as the answer is $-1$. Where did I mess up? Thanks.

Juniven Acapulco
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alan
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4 Answers4

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Notice that since $x$ is tending to $-\infty$ then $x=-\sqrt{x^{2}}$. So when you let $x$ enter the square root in the denominator the minus that was originally there becomes a plus due to cancellation. That is:

$$\frac{-2}{\frac{x-\sqrt{x^{2}+2x}}{x}}=\frac{-2}{1-\frac{1}{x}\sqrt{x^{2}+2x}}=\frac{-2}{1+\sqrt{1+\frac{2}{x}}}$$

which tends to $-1$ as $x$ tends to $-\infty$.

user71352
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For better clarity set $\dfrac1x=-y\implies y\to0^+$

$$\lim_{x\to-\infty}(x+\sqrt{x^2+2x})=\lim_{y\to0^+}\frac{\sqrt{1-2y}-1}y$$

$$=\lim_{y\to0^+}\frac{1-2y-1}{y(\sqrt{1-2y}+1)}$$ Hope you can take it home from here

lab bhattacharjee
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Hint :

Set $x=-y$, then we will have $$ \lim_{y\to\infty} \left(\sqrt{y^2 - 2y}-y\right) $$ Now, repeat the process as you have done (multiplying by its conjugate and dividing by the highest power of denominator).

Tunk-Fey
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Me I think, negative infinity minus negative infinity is not indeterminate, it has a meaning, therefore the limit does not exist. You only multiply with the conjugate if the limit are indeterminate

Abel
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